Question: Solve for $x$ : $6x^2 + 102x + 420 = 0$
Explanation: Dividing both sides by $6$ gives: $ x^2 + {17}x + {70} = 0 $ The coefficient on the $x$ term is $17$ and the constant term is $70$ , so we need to find two numbers that add up to $17$ and multiply to $70$ The two numbers $10$ and $7$ satisfy both conditions: $ {10} + {7} = {17} $ $ {10} \times {7} = {70} $ $(x + {10}) (x + {7}) = 0$ Since the following equation is true we know that one or both quantities must equal zero. $(x + 10) (x + 7) = 0$ $x + 10 = 0$ or $x + 7 = 0$ Thus, $x = -10$ and $x = -7$ are the solutions.